The correct option is C 62729 = 9 x 9x 9 ∴ √(729)= √(9× 9× 9) nbsp; = 9 nbsp; 3^3 = 3× 3× 3 = 27 So, √(729) +3^3 = 9+27 ...

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Cubes Relation with Cube Numbers

√729+33=

Question

$\sqrt[3]{729} + (3)^{3}$ = ________

6^{3}

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2^{5}

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6^{2}

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2^{6}

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Solution

The correct option is C $6^{2}$
729 = 9 x 9x 9
$∴$ $\sqrt[3]{729} = \sqrt[3]{9 \times 9 \times 9} = 9$

$3^{3} = 3 \times 3 \times 3 = 27$

So, $\sqrt[3]{729} + 3^{3} = 9 + 27 = 36 = 6^{2}$

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Similar questions

\frac{1}{{(243)}^{x}} = {(729)}^{y} = 3^{3}

$2$	$3$	$4$	$- 5$	__	$8$	__
$2^{3} = 8$	$3^{3} = . . . . . .$	$. . . . = 64$	$. . . . . = . . . . .$	$6^{3} = . . . . .$	$. . . . . = . . . .$	$. . . . = - 729$

\sqrt[3]{729} + \sqrt{729}

(?)^{3} = 729

\sqrt[3]{\frac{x}{729}} + \sqrt[3]{\frac{8 x}{729}} + \sqrt[3]{\frac{27 x}{5832}} = 1

x

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